If f ' x = tan - 1 ⁡ sec ⁡ x + tan ⁡ x , - π 2 < x < π 2 and f 0 = 0 , then f 1 is equal to:

Question

If f ' x = tan - 1 &#8289; sec &#8289; x + tan &#8289; x , - &#960; 2 &#60; x &#60; &#960; 2 and f 0 = 0 , then f 1 is equal to:

Aicharya · Accepted Answer

f ' x = tan - 1 &#8289; sec &#8289; x + tan &#8289; x = tan - 1 &#8289; 1 + sin &#8289; x cos &#8289; x = tan - 1 &#8289; 1 - cos &#8289; &#960; 2 + x sin &#8289; &#960; 2 + x &#8658; f ' ( x ) = tan - 1 &#8289; 2 sin 2 &#8289; &#960; 4 + x 2 2 sin &#8289; &#960; 4 + x 2 cos &#8289; &#960; 4 + x 2 &#8658; f ' x d x = &#960; 4 + x 2 d x &#8658; f x = &#960; 4 x + x 2 4 + c &#8757; f 0 = 0 &#8658; c = 0 So, f 1 = &#960; + 1 4

If f ' x = tan - 1 ⁡ sec ⁡ x + tan ⁡ x , - π 2 < x < π 2 and f 0 = 0…

Solution

If f &#39; x = tan - 1 &#8289; sec &#8289; x + tan &#8289; x , - &#960; 2 &#60; x &#60; &#960; 2 and f 0 = 0…

Solution

If f ' x = tan - 1 ⁡ sec ⁡ x + tan ⁡ x , - π 2 < x < π 2 and f 0 = 0…